Given E deg(Fe^3+/Fe^2+) = 0.77 V and E deg(Fe^2+/Fe) = -0.44 V at 298 K. For the disproportionation 3Fe^2+ -> 2Fe^3+ + Fe, the equilibrium constant is of the order of (take (2.303RT)/(F) = 0.0591 V)

10^-41; Fe^2+ is stable to disproportionation

10^-20; Fe^2+ is stable to disproportionation

10^41; Fe^2+ disproportionates completely

10^-61; Fe^2+ is stable to disproportionation

JEE Advanced levelSingle correct optionDifficulty 3 of 3ec069

Given $E^{\circ} (Fe X^{3 +} / Fe X^{2 +}) = 0.77 V$ and $E^{\circ} (Fe X^{2 +} / Fe) = - 0.44 V$ at 298 K. For the disproportionation $3 Fe X^{2 +} 2 Fe X^{3 +} + Fe$ , the equilibrium constant is of the order of (take $\frac{2.303 RT}{F} = 0.0591 V$ )

In the disproportionation

Fe X^{2 +}

is reduced to Fe at the cathode (

- 0.44 V

) and oxidised to

Fe X^{3 +}

at the anode (

0.77 V

E^{\circ} = - 0.44 - 0.77 = - 1.21 V

. The balanced equation transfers

n = 2

electrons.

lo g K = \frac{2 \times ( - 1.21 )}{0.0591} = - 40.9

, so

K \approx 1 0^{- 41}

Fe X^{2 +}

does not disproportionate. Option B uses

n = 1

; option D uses

n = 3